What is the solution for y^{''}-2y^'+y=xe^x ?

The solution for y^{''}-2y^'+y=xe^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^3)/6

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y^{''}-2y^'+y=xe^x

y^{^{''}} - 2 y^{^{'}} + y = x e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = x e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6}

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{t} sin (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{e ^{- 2 x}}{x ^{2}}

y^{^{''}} - 5 y^{^{'}} = 2 x^{3} - 4 x^{2} - x + 6

(D^{2} + 4 D + 5) y = 50 x + 13 e^{3 x}

What is the solution for y^{''}-2y^'+y=xe^x ?
The solution for y^{''}-2y^'+y=xe^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^3)/6